Interest rates. What is the difference between $I=I_0(1+r)^t$ and $\frac{dI}{dt}=rI$?

Question

When I was in school, we used this method for generating the amount of money would be in a bank account after $t$ years with interest rate $r$: $$I=I_0(1+r)^t \text{ where }I_0\text{ is the initial investment}$$ but now in my mathematical finance university course, i'm taught that that it is worked out through using: $$\frac{dI}{dt}=rI \Rightarrow I(t)=I_0e^{rt}$$ I've noticed that these are not equivalent, so which is correct? I'm assuming their both correct for different methods of working out interest but can someone explain the differences?

Answer 1

Compare $$I_{n+1}-I_n=rI_n\implies I_n=(1+r)^nI_0\qquad (n\in\mathbb N_0),$$ and $$I'(t)=rI(t)\implies I(t)=\mathrm e^{rt}I(0)\qquad (t\in\mathbb R_+).$$

Answer 2

One of them is in continuous time and the other in discrete time. In one case the amount is multiplied by $1+r$ every time $1$ unit of time passes; in the other case it is multiplied by $e^r$, so $e^r-1$ rather than $r$ is the effective rate. The nominal rate in the latter case is $r$, and that means the effective rate can be made as close to $r$ as desired by making the amount of time small enough. Say in a millionth of a second the balance grows by a certain amount. If it continued growing at that rate in dollars per year, the balance might reach $1+r$ times the initial balance after one year passes. But the rate of growth doesn't continue to be the same, but increases with the balance, so it ends up more than $1+r$ times the original amount. That's the difference between nominal and effective interest rates if compounding is done every millionth of a second. That is almost, but not quite, the same as continuous compounding.